How many ways are there to distribute $18$ distinguisable object into
a-) $5$ distinguishable boxes so that the boxes have $1,2,4,5,6$ objects in them, respectively.
b-) $5$ distinguishable boxes so that the boxes have $1,2,4,5,6$
objects in them.c-) $5$ distinguishable boxes so that the boxes have $4,2,2,5,5$
objects in them, respectively.d-) $5$ distinguishable boxes so that the boxes have $4,2,2,5,5$
objects in them.e-) $5$ indistinguishable boxes so that the boxes have $1,2,4,5,6$
objects in them, respectively.f-) $5$ indistinguishable boxes so that the boxes have $1,2,4,5,6$
objects in them.g-) $5$ indistinguishable boxes so that the boxes have $4,2,2,5,5$
objects in them, respectively.h-) $5$ indistinguishable boxes so that the boxes have $4,2,2,5,5$
objects in them.
My attempt:
a-) $ C(18,1) \times C(17,2) \times C(15,4) \times C(11,5) \times C(6,6) $
b-) $ 5! \times (C(18,1) \times C(17,2) \times C(15,4) \times C(11,5) \times C(6,6)) $
c-) $C(18,4) \times C(14,2) \times C(12,2) \times C(10,5) \times C(5,5) $
d-) $(\frac{5!}{2!\times 2! \times 1!})\times(C(18,4) \times C(14,2) \times C(12,2) \times C(10,5) \times C(5,5)) $
e-) I think this question is invalid , because if the boxes are indistinguishable then we cannot talk about respectivity.
f-) In this question , i firstly thought them like distinguishable boxes, after that i divide them by
$5!$ to obtain indistinguishable form such that $\frac{(C(18,1) \times C(17,2) \times C(15,4) \times C(11,5) \times C(6,6))}{5!} $
g-)I think this question is invalid, because if the boxes are indistinguishable then we cannot talk about respectivity.
h-) In this question, i firstly thought them like distinguishable boxes, after that i divide them by
$\frac{5!}{2!\times 2! \times 1!}$ such that $\frac{C(18,4) \times C(14,2) \times C(12,2) \times C(10,5) \times C(5,5)}{\frac{5!}{2!\times 2! \times 1!}} $
Is my solutions correct? If not, can you correct me, please ?
Best Answer
I only have comments on your working for (f) and (h).
(f) The idea is to create $5$ indistinguishable heaps of $1, 2, 4, 5$ and $6$ objects and that is simply $C(18,1) \times C(17,2) \times C(15,4) \times C(11,5) \times C(6,6)$, same as in your answer $(a)$.
(h) It is similar to (f) but we have heaps of $4, 2, 2, 5$ and $5$ objects. Now note that $C(18,4) \times C(14,2) \times C(12,2) \times C(10,5) \times C(5,5)$ will order two heaps of $2$ objects and will order two heaps of $5$ objects but we do not want that.
So the answer to this is $ \ \displaystyle \frac{1}{2 \cdot 2} \cdot C(18,4) \times C(14,2) \times C(12,2) \times C(10,5) \times C(5,5)$